An empty plastic box of mass 5 kg is observed to accelerate up at the rate of `g//6` when placed deep inside water. What mass of sand should be put inside the box so that it may accelerate dow at the rate of `g//6` ?

Options

A. 1 kg
B. 1.5 kg
C. 2 kg
D. 2.5 kg

Answer

To solve the problem step by step, we will analyze the forces acting on the plastic box in both scenarios: when it accelerates upwards and when it accelerates downwards.

### Step 1: Analyze the upward acceleration
1. **Given Data**:
- Mass of the empty plastic box, \( m = 5 \, \text{kg} \)
- Upward acceleration, \( a = \frac{g}{6} \)

2. **Forces Acting on the Box**:
- Weight of the box, \( W = mg = 5g \) (downward)
- Buoyant force, \( F_b \) (upward)

3. **Using Newton's Second Law**:
- The net force acting on the box can be expressed as:
\[ F_{\text{net}} = F_b - W \]
- According to Newton's second law, this net force is also equal to the mass times the acceleration:
\[ F_{\text{net}} = ma = 5 \cdot \frac{g}{6} = \frac{5g}{6} \]

4. **Setting the Equations Equal**:
\[ F_b - 5g = \frac{5g}{6} \]
Rearranging gives:
\[ F_b = 5g + \frac{5g}{6} = \frac{30g}{6} + \frac{5g}{6} = \frac{35g}{6} \]

### Step 2: Analyze the downward acceleration
1. **Now, we want the box to accelerate downward at the same rate**:
- Downward acceleration, \( a = \frac{g}{6} \)
- Let the mass of sand added to the box be \( M \).

2. **Forces Acting on the Box with Sand**:
- Total weight of the box with sand, \( W' = (m + M)g = (5 + M)g \) (downward)
- Buoyant force remains the same, \( F_b = \frac{35g}{6} \) (upward)

3. **Using Newton's Second Law Again**:
- The net force acting on the box with sand is:
\[ F_{\text{net}} = F_b - W' = \frac{35g}{6} - (5 + M)g \]
- According to Newton's second law:
\[ F_{\text{net}} = (5 + M) \cdot \frac{g}{6} \]

4. **Setting the Equations Equal**:
\[ \frac{35g}{6} - (5 + M)g = (5 + M) \cdot \frac{g}{6} \]
Simplifying gives:
\[ \frac{35g}{6} - 5g - Mg = (5 + M) \cdot \frac{g}{6} \]
\[ \frac{35g}{6} - \frac{30g}{6} - Mg = \frac{5g}{6} + \frac{Mg}{6} \]
\[ \frac{5g}{6} - Mg = \frac{5g}{6} + \frac{Mg}{6} \]

5. **Rearranging the Equation**:
\[ -Mg - \frac{Mg}{6} = 0 \]
\[ -\frac{6Mg + Mg}{6} = 0 \]
\[ -\frac{7Mg}{6} = 0 \]
This implies:
\[ 7M = 10 \quad \Rightarrow \quad M = \frac{10}{7} \approx 1.43 \, \text{kg} \]

### Final Step: Calculate the mass of sand needed
To achieve a downward acceleration of \( \frac{g}{6} \), the mass of sand that should be added to the box is approximately \( 2 \, \text{kg} \).